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So I recently came upon the following complex algebra problem:

$$ z^{\frac{4}{3}} = -2 $$

So, to solve it I have to find the z values that solve the following:

$$ z = (-2)^{\frac{3}{4}} $$

To do this I express -2 in exponential form:

$$ z = (2e^{i(\pi + 2\pi n)})^{\frac{3}{4}} $$

Then, I solve for that trying for $n=0,1,2,3$ and I come up with 4 roots: $$ z_1 = 2^{\frac{3}{4}}e^{i\frac{\pi}{4}} $$ $$ z_2 = 2^{\frac{3}{4}}e^{i\frac{3\pi}{4}} $$ $$ z_3 = 2^{\frac{3}{4}}e^{i\frac{5\pi}{4}} $$ $$ z_4 = 2^{\frac{3}{4}}e^{i\frac{7\pi}{4}} $$

However, if I try to check these solutions for the original problem, only $z_2$ and $z_3$ succeed, while $z_1$ and $z_4$ do not solve the initial equation. Even plugging the original equation into Wolfram, gives me just those two roots.

I have been thinking about this over and over and don't understand where I'm going wrong or what is it that I'm failing to consider. Does anybody have any idea of where I'm going wrong?

Thank you in advance

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  • $\begingroup$ Btw, Mathjax works in titles too $\endgroup$ – sai-kartik Jun 23 '20 at 2:57
  • $\begingroup$ Thanks! I didn't know that $\endgroup$ – Alejandro Flores Jun 23 '20 at 2:59
  • $\begingroup$ I would have said to keep the roots with principle arguements but looks like it doesn't match with what you're getting.. $\endgroup$ – sai-kartik Jun 23 '20 at 3:02
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    $\begingroup$ From this chatroom,you can see that -π to π is the definition of the principle argument, and it helps to keep it in this range, cus then you're gonna easily avoid repeated solutions $\endgroup$ – sai-kartik Jun 23 '20 at 3:59
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    $\begingroup$ Also you can convert 5π/4 and 7π/4 further to make it fall in the principle argument range $\endgroup$ – sai-kartik Jun 23 '20 at 4:00
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I presume you're treating $z^{4/3}$ as a multivalued function, and you're allowing any $z$ such that any branch of $z^{4/3}$ is $2$. By definition, $z^{4/3} = \exp((4/3) \log(z))$ where $\log(z)$ is any branch of the logarithm of $z$. If $\text{Log}(z)$ is the principal branch (with imaginary part in $(-\pi, \pi]$), the other branches of $\log(z)$ are $\text{Log}(z) + 2 \pi i n$ for arbitrary integers $n$, and the corresponding branches of $z^{4/3}$ are $\exp((4/3) \text{Log}(z) + (8 \pi i n/3))$. There are three possibilities, corresponding to the values of $n \mod 3$. Now this is supposed to be $-2 = 2 \exp(\pi i)$. Thus for $n \equiv 0 \mod 3$, $$2 = \exp((4/3) \text{Log}(z) - \pi i)$$ where $\text{Im}((4/3) \text{Log}(z) - \pi i) = 0$ and $\text{Re}((4/3) \text{Log}(z) = \text{Log}(2)$. We get either $\text{Log}(z) = (3/4) \text{Log}(2) + 3 \pi i/4$, i.e. $z = 2^{3/4} e^{3 \pi i/4}$, or $\text{Log}(z) = (3/4) \text{Log}(2) - 3 \pi i/4$, i.e. $z = 2^{3/4} e^{-3\pi i/4}$.

(this $2^{3/4}$ being the real $3/4$ power).

For $n \equiv 1 \mod 3$, $$2 = \exp((4/3) \text{Log}(z) + 5 \pi i/3)= \exp((4/3) \text{Log}(z) - \pi i/3$$ where $\text{Im}((4/3) \text{Log}(z) - \pi i/3 = 0$. We get $\text{Log}(z) = (3/4) \text{Log}(2) + \pi i/4$, or $z = 2^{3/4} e^{\pi i/4}$.

For $n \equiv 2 \mod 3$, $$2 = \exp((4/3) \text{Log}(z) + 13 \pi i/3) = \exp((4/3) \text{Log}(z) + \pi i/3$$ where $\text{Im}((4/3) \text{Log}(z) + \pi i/3 = 0$. We get $\text{Log}(z) = (3/4) \text{Log}(2) - \pi i/4$, or $z = 2^{3/4} e^{-\pi i/4}$.

So there are indeed four solutions. However, if you try to verify these with Mathematica or most other computer algebra systems, they won't all work, as they like to use the principal branch rather than multivalued functions.

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  • $\begingroup$ If $\operatorname {Log}$ is the principal branch, then $$\operatorname {Re}(4 \operatorname {Log}(z)/3) = \operatorname {Log} 2 \land \operatorname {Im}(4 \operatorname {Log}(z)/3 - \pi i) = 0$$ (the case $n \equiv 0 \bmod 3$) has only one solution. The second equation should be $$\operatorname {Im}(4 \operatorname {Log}(z)/3 - \pi i) = 2 \pi k.$$ $\endgroup$ – Maxim Jul 12 '20 at 18:33
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Please note that the complex function $f(z)=z^{\frac{1}{n}}$, $n \in \mathbb{N}, \, n \ge 2$ is a multivalued function. Writing the function in polar form,$$z=re^{i \theta } \qquad \rightarrow \qquad f(z)=(re^{i \theta })^{\frac{1}{n}}=r^{\frac{1}{n}}e^{i \frac{\theta }{n}},$$we can easily conclude that a point with arguments $\theta$, $\theta + 2\pi$, ..., $\theta + 2(n-1)\pi$ in the domain plane corresponds to $n$ distinct points with the arguments $\frac{\theta }{n}$, $\frac{\theta }{n}+\frac{2\pi }{n}$, ..., $\frac{\theta }{n}+\frac{2(n-1)\pi }{n}$ in the image plane. In other words, this function is a one-to-$\bf{n}$ correspondence.

With a similar argument, one can show that the function $f(z)=z^{\frac{4}{3}}$ is a one-to-three correspondence.

You have solved $z^{\frac{4}{3}}=-2$ correctly. However, please note that to check the solutions for the original problem, you should use the same representation of points you reached by solving the problem, that is,$$z_1 = 2^{\frac{3}{4}}e^{i\frac{3\pi}{4}} $$$$z_2 = 2^{\frac{3}{4}}e^{i\frac{9\pi}{4}}$$$$z_3 =2^{\frac{3}{4}}e^{i\frac{15\pi}{4}}$$$$z_4 =2^{\frac{3}{4}}e^{i\frac{21\pi}{4}},$$which are clearly satisfy the original problem. Otherwise, you may get other values of $z^{\frac{4}{3}}$ not satisfying the original problem.

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